Series 215 problem 05.mws

1. More estimating the remainder.

Integrals can be used to estimate the remainder of a convergent series. For instance, suppose I wish to evaluate the sum of the p-series, sum(1/(n^2),n = 1 .. infinity) . Euler claims that this sums to Pi^2/6 We compare the series to f(x) = 1/(x^2) , noting that f is continuous, positive and decreasing on [ 1, infinity ) and use the remainder estimate to determine a value of n so that our sum is within .0001 of the actual sum.

> f:=x->1/x^2;

f := proc (x) options operator, arrow; 1/(x^2) end ...

> int(f(x),x=10001..infinity);

1/10001

> evalf(sum(1/n^2,n=1..10000));

1.644834073

> evalf(Pi^2/6);

1.644934068

Notice that the two numbers agree within the error of our estimate.

Submission:

(a) Find the partial sum S[10] of the series sum(1/(n^4),n = 1 .. infinity) . Estimate the error in using S[10] as an

approximation to the sum of the series.

(b) Use the fact that

S[n]+Int(f(x),x = n+1 .. infinity) <= S and S <= S[n] + Int(f(x),x = n .. infinity)

with n = 10 to give an improved estimate of the sum.

(c) Find a value of n so that S[n] is within 0.00001 of the true sum.

(d) How many terms of the series sum(1/(n*ln(n)^2),n = 2 .. infinity) would you need to add to find its sum to within 0.01 ?

Submission worksheet:

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2. Estimating the sum of a positive series using the integral test.

Let Sum(a[n],n = 1 .. infinity) be a positive series and suppose that f(n) = a[n] for some function f defined on [ N, infinity ) which is continuous, positive and decreasing on that interval. Then we know that the series converges precisely when the integral Int(f(x),x = N .. infinity) converges. Suppose that the series does converge and that Sum(a[n],n = 1 .. infinity) = A . Let R[N] = A-s[n] . Then R[N] = Sum(a[n],n = N+1 .. infinity) , and we can show using a picture that

Int(f(x),x = N+1 .. infinity) <= R[N] <= Int(f(x),x = N .. infinity)

This formula allows us to determine bounds on the error in using s[n] to estimate the value A to which the series converges.

Let us consider the p -series Sum(1/(n^2),n = 1 .. infinity) , which the mathematician Euler discovered converges to Pi^2/6 . We can compare this series to the function f(x) = 1/(x^2) .

It is easy to see that this function is positive, continuous and decreasing on [ 1, infinity ). Let us compute the integral

Int(1/(x^2),x = a .. infinity) = Limit(Int(1/(x^2),x = a .. b),b = infinity) = Limit(1/a-1/b,b = infinity) = 1/a

From this it follows that 1/(N+1) <= R[N] <= 1/N , Thus if we want to estimate the sum of the series to within .001, we need to use 1000 terms in the sequence. Let us compare our estimate with Euler's exact answer.

> est:=evalf(sum(1/n^2,n=1..1000));

est := 1.643934568

> ans:=evalf(Pi^2/6);

ans := 1.644934068

> ans-est;

.999500e-3

We could also improve our estimate slightly by using the fact that we know that our error is at least 1/1001, so we could refine the estimate to be

> est:=evalf(sum(1/n^2,n=1..1000)+1/1001);

est := 1.644933569

> ans-est;

.499e-6

Notice the big improvement in the estimate we just obtained. This suggests that if we use s[n]+Int(f(x),x = N+1 .. infinity) to approximate the sum of the series, we obtain a better estimate of its sum.

Submission:

The Riemann Zeta Function plays an important role in mathematics. The question of where its zeros are located is the subject of the Riemann Hypothesis, an important unsolved conjecture. The value of the Riemann Zeta function is given by Zeta(p) = sum(1/(n^p),n = 1 .. infinity) when 1 < p . The objective of this exercise is to estimate the value of the Riemann Zeta function at p = 3 using the integral test, in other words, to estimate the sum of the p- series for p = 3 .

a) Define the term a[n] for this series in Maple, and then compute the sum of the series using Maple. Finally, get a floating point estimate of this value.

b) Find the function f(x) which we use in the integral test, and show it satisfies the hypotheses of the integral test.

c) Find a number N which is such that if N < n , the error in using the value s[n] to approximate the value of the series is no worse than .0001.

d) Determine the error in using s[n]+Int(f(x),x = N+1 .. infinity) to approximate the sum of the series.

Submission worksheet:

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